Question 796ac

May 12, 2015

Formality is just a way to express the concentration of a substance in solution, regardless of how that substance exists in that solution.

Formality and molarity are the same thing for compounds that do not dissociate when dissolved in a solvent. However, they are different for compounds that do dissociate when dissolved.

In your case, you'd calculate formality just like you would molarity. Use potassium nitrate's molar mass to determine how many moles you're dealing with

2.02cancel("g") * ("1 mole "KNO_3)/(101.1cancel("g")) = "0.01998 moles" $K N {O}_{3}$

The solution's formality will be

$F = \frac{n}{V} = \text{0.01998 moles"/(600 * 10^(-3)"L") = "0.03 F}$

This is exactly how molarity is calculated. However, you wouldn't say that your potassium nitrate solution has a molarity of 0.03 M because you don't actually have potassium nitrate in that solution.

What you have are potassium cations ,${K}^{+}$, and nitrate anions, NO_3""^(-). So your solution would be 0.03 M in ${K}^{+}$ and 0.03 M in NO_3""^(-)#.

So, when you say you have a 1 M solution of sodium chloride, you actually have a 1 F solution of $N a C l$ and a 1 M solution in $N {a}^{+}$ and in $C {l}^{-}$.

By comparison, if you have a 1 F glucose solution, you also have a 1 M glucose solution, because glucose does not dissociate in aqueous solution.